题意
description
You are given an array a1,a2,...,an(∀i∈[1,n],1≤ai≤n). Initially, each element of the array is **unique**. Moreover, there are m instructions. Each instruction is in one of the following two formats: 1. (1,pos),indicating to change the value of apos to apos+10,000,000; 2. (2,r,k),indicating to ask the minimum value which is **not equal** to any ai ( 1≤i≤r ) and **not less ** than k. Please print all results of the instructions in format 2.
input
The first line of the input contains an integer T(1≤T≤10), denoting the number of test cases. In each test case, there are two integers n(1≤n≤100,000),m(1≤m≤100,000) in the first line, denoting the size of array a and the number of instructions. In the second line, there are n distinct integers a1,a2,...,an (∀i∈[1,n],1≤ai≤n),denoting the array. For the following m lines, each line is of format (1,t1) or (2,t2,t3). The parameters of each instruction are generated by such way : For instructions in format 1 , we defined pos=t1⊕LastAns . (It is promised that 1≤pos≤n) For instructions in format 2 , we defined r=t2⊕LastAns,k=t3⊕LastAns. (It is promised that 1≤r≤n,1≤k≤n ) (Note that ⊕ means the bitwise XOR operator. ) Before the first instruction of each test case, LastAns is equal to 0 .After each instruction in format 2, LastAns will be changed to the result of that instruction. (∑n≤510,000,∑m≤510,000 )
分析
题目中，共有两种操作，一种是将第k个数字+1000,0000，一种是求一个最小的、不等于ai(1 <= i <= r)的，且不小于k的数字。
注意 这个数字，可以不出现在array中
因为array中的数字各不相同，因此，不等于ai(1 <= i <= r)，等价于 这个数字可能会出现在 ai(r + 1 <= i <= n)。
为什么是可能？因为由于第一种操作+1000,10000，会使得进行第一种操作的那个数字，无论如何都不会和ai重复（1000,0000 比 51,0000大太多了）。
所以，这个题，我们可以将主席树进行适当的修改，每次查找ai[r + 1, n]中，不小于k的最小数字。
由于本题中，array由1-n的n个数字组成，因此不用考虑数字的去重(unique)，以及 离散化问题。
除了上面两种情况，还有一种情况，我们没考虑到，如果k = n (已经过异或运算)，那么这个答案还有可能是 n + 1。
综上，答案一共有三种情况：
- ai[r + 1, n] 中的某个数字 （可以用主席树求解）
- 进行过第一种操作的某个数字 （将进行过第一种操作的数字放入set中）
- n + 1
三个数字，求最小值即可
代码
#include <iostream>
#include <cstdio>
#include <set>
#include <algorithm>
int const maxn = 530000;
int const inf = 0x3f3f3f3f;
using namespace std;
int a[maxn], b[maxn];
int root[maxn << 5];//第几个版本的根节点编号
int lc[maxn << 5], rc[maxn << 5], sum[maxn << 5];
int sz;//节点个数
int n, m;
void build(int &rt, int l, int r) {
rt = ++sz;
if (l == r) return;
int mid = (l + r) >> 1;
build(lc[rt], l, mid);
build(rc[rt], mid + 1, r);
}
int update(int id, int l, int r, int pos) {
int _id = ++sz;
lc[_id] = lc[id], rc[_id] = rc[id], sum[_id] = sum[id] + 1;
if (l == r) return _id;
int mid = (r + l) >> 1;
if (pos <= mid)
lc[_id] = update(lc[id], l, mid, pos);
else
rc[_id] = update(rc[id], mid + 1, r, pos);
return _id;
}
//查询 不比k大的最小数字
int query(int p, int q, int l, int r, int k) {
if (l == r) return l;
int x1 = sum[lc[q]] - sum[lc[p]];
int x2 = sum[rc[q]] - sum[rc[p]];
int mid = (l + r) >> 1;
int ans = inf;
if (x1 > 0 && mid >= k)
ans = query(lc[p], lc[q], l, mid, k);
//这个if不能写为else，因为第一个if可能无法得到结果，返回inf
if(ans == inf && x2 > 0 && r >= k)
ans = query(rc[p], rc[q], mid + 1, r, k);
return ans;
}
int main() {
int T;
scanf("%d", &T);
while (T--) {
while (~scanf("%d %d", &n, &m)) {
sz = 0;
set<int> s;
int lastAns = 0;
for (int i = 1; i <= n; i++) {
scanf("%d", &a[i]);
b[i] = a[i];
}
sort(b + 1, b + n + 1);
build(root[0], 1, n);
for (int i = 1; i <= n; i++) {
int pos = lower_bound(b + 1, b + n + 1, a[i]) - b;
root[i] = update(root[i - 1], 1, n, pos);
}
while (m--) {
int l;
scanf("%d", &l);
if (l == 1) {
int pos = 1; // 随意初始化
scanf("%d", &pos);
pos ^= lastAns;
//cout << "pos = " << pos << endl;
s.insert(a[pos]);
}
else {
int l, k;
scanf("%d %d", &l, &k);
l ^= lastAns;
k ^= lastAns;
//cout << "l = " << l << "k = " << k << endl;
int ansPos = query(root[l - 1 + 1], root[n], 1, n, k);
lastAns = (ansPos == inf) ? inf : b[ansPos];
set<int>::iterator it = s.lower_bound(k);
//
if (it != s.end()) lastAns = min(lastAns, *it);
lastAns = min(lastAns, n + 1);
printf("%d\n", lastAns);
}
}
}
}
return 0;